### Discussion :: Numbers - General Questions (Q.No.32)

Ashok Kumar said: (Jan 27, 2011) | |

n=[(l-a)/d]+1 n=150 |

Vivek said: (Apr 29, 2011) | |

Adding the given three deghit it means =150 1+5+0/6=remander is 0 Hence if the adding number is not same in divedent and remander is not 0. |

Mahesh Patil said: (Aug 1, 2011) | |

Total Three digit numbers = 999-100+1 = 900 so numbers Divisible by 6 = 900 / 6 = 150 (Only Quotient ... don't consider the remainder) |

Rajkin said: (Nov 12, 2011) | |

Thanks mahesh patel. |

Sworna said: (Feb 25, 2012) | |

@Mahesh. Why it is 999-100+1? I can't understand please help me. |

Neha Nagar said: (Mar 12, 2012) | |

@sworna Dear sworna here +1 is because 100 is also included in three digit no we can not leave 100 here. So it will b better to solve like this....3 digit nos-2 digit nos. So that 999-99=900. Hope you have got it. |

Gis said: (Jun 20, 2012) | |

@sworna 999-100+1 is taken based on arithmetic progression . last 3-digit no: is 999 and first 3-digit no: is 100 . according to the equation n=(l-a)/d +1 here l=999,a=100, d=1 so n=(999-100)/1 +1=900 hope u can understand |

Lalit said: (Jun 23, 2012) | |

Another short technique is last 3 digit no is 999 so we divide 999/6= 166 we don't consider decimal part. Now, we only need 3 digit number so 100/6= 16 therefore, 166-16=150 answer. 166 is the total number divisible by 6. We subtracted 100 because before 100 there is only 2 digit number divide by 6. |

Ashish said: (Feb 5, 2013) | |

If we want to find how many numbers present in 3 digit no which is divisible by any number. Then here is method. Eg: How many 3-digit numbers are completely divisible 8 ? Answer: 1000/8=125. 100/8=12 with remainder 4. 125-12=113. So 113 numbers are divisible from 100 to 999 by 8. |

Harpreet said: (May 10, 2013) | |

@Ashish total numbers divisible by 8 comes 112 & not 113. |

Udaya Santhi said: (Aug 20, 2013) | |

6 written as 2*3. The number completely divisible by both numbers 2&3. If the number is divisible by 2 that number should be an even. If the number is divisible by 3 then taken sum of digits. Option verification: Only two even numbers are there 150 & 166. These two numbers are divisible by 2. 1+5+0 = 6(which is divisible by 3), 1+6+6 = 13(which is not divisible by 3). Hence 150 is the answer. |

Swathi said: (Feb 28, 2014) | |

149:1+4+9 = 16 it is not divisible by 6. 150:1+5+0 = 6 it is divisible by 6. 151:1+5+1 = 7 not divisible by 6. 166:1+6+6 = 13 not divisible by 6. Ans :150 (B). |

Ishan said: (Aug 1, 2014) | |

Directly, on dividing 150 by 6 gives o as a reminder and as a quotient it gives 25 so that 150 comes in the multiple of 25 in the 6 place. 150/6 = 25. 25*6 = 150. |

Samir said: (Aug 19, 2014) | |

It can take any no.of digit so, Last digit - first digit(996-102) = 894. Divide by divisible no. 894/6 = 149. In formula opposite (n-1) +1. (149+1) = 150. |

Ayushi said: (Jun 27, 2015) | |

Divisibility rule of 6. A no. is divisible by 6 if the no. is even and sum of digit is divisible by 3. |

Joshua said: (Jul 13, 2015) | |

What if you try the formula for geometric sequence which is an = a1 x r^n-1. |

Vicky Kumar Verma said: (Jan 20, 2016) | |

999/6 = 166 ignore remainder. 99/6 = 16 ignore remainder. Now, 166-16 = 150 answer. |

Rahul said: (Jul 29, 2016) | |

Please give me a solution for "What is the sum of 3 digits number which is completely divisible by 3, then find the sum". |

Samir Das said: (Aug 23, 2016) | |

Thanks for the answer. It's useful to me. |

Rohit said: (Dec 19, 2016) | |

100 - 999 these are three digits no, (last term - first term) + 1 = 900. And divide 900 by 6 i. e. we obtained 150 is final answer. |

Shradha Khirid said: (Jun 28, 2017) | |

If no is divisible by 3 and 2 then it is easily divisible by 6. |

Tanu said: (Aug 28, 2017) | |

Greatest 3 digit number that is divisible by 6 is 996 (q=166). The smallest is 102(q=17). Apply simple number rule to count number of values between 166 to 17, i.e. 166-17+1=150. Thus this gives you the number of 3 digits divisible by 6. |

Damini said: (Sep 16, 2017) | |

Write numbers from sequence 149 to 387 like 149150159. How many time does 1 occur? |

Vatsal said: (Dec 5, 2017) | |

Number is divided by 6 only when it is perfectly divided by 2 and 3 both. So Only 150 is divided by both 2 and 3. |

Nithya said: (May 25, 2018) | |

Divisiablity shortcut for 6 = the number should divide be both 2 & 3. a. 149 / 2 = no because unit digit is odd. b. 150 /2 = yes because unit digit ends in 0, then check for 3 -> 150 / 3 = 1+5+0 = 6 = 6/3 = 2, so answer is B. |

Tushar said: (Jul 9, 2018) | |

Step 1) Consider highest 3 digit number divisible by 3 which is 996, dividing it by 6 we get remainder as 166. Since we want only 3 digit numbers which are divisible by 3 we need to subtract the 2 digit numbers in here. Step 2) Consider highest 2 digit number divisible by 3 which is 96,dividing it by 6 we get remainder as 16.That means there are 16 (2 digit numbers) divisible by 6. Step 3) Now subtract these two corresponding remainders we get 166-16 = 150. |

Tirtha said: (Nov 12, 2018) | |

By taking the multiples of 2*3 for 6 we can operate the divisibility rule for this question. As for 2, the ending digit should be 0, 2, 4, 6, 8 and for 3 sum of all the digit should be divisible by 3. So, 150 is divisible by both 2 & 3. This is the correct answer. |

Mahesh said: (Aug 20, 2019) | |

6 = 2 * 3 which are co-actors, then the number divisible by these two will also divisible by 6. |

Amar said: (Sep 2, 2020) | |

Lowest 3 digit number which is divisible by 3 is 102. Highest 3 digit number which is divisible by 3 is 999. So as per the formula a+(n-1)d=l. The answer will be 300. |

Sahul Hameed said: (Apr 21, 2021) | |

As per the Divisibility rule; If last digit is even it will divided by 2 and 3 also divided by 6. In this problem only 166 can sacrifice this method. So the answer is 166. |

Anand said: (Jun 27, 2021) | |

Let the highest 3 digit num divided by 6 is 996, then 996/6= 166. Let the highest 2 digit num divided by 6 is 96, then 96/6= 16, Now 166-16= 150. |

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